DOE HDBK DOE Fundamentals Handbook Electrical Science

					Basic AC Power                                                           THREE-PHASE CIRCUITS



                            THREE-PHASE CIRCUITS

         The design of three-phase AC circuits lends itself to a more efficient method of
         producing and utilizing an AC voltage.

         EO 1.5        STATE the reasons that three-phase power systems are
                       used in the industry.

         EO 1.6        Given values for current, voltage, and power factor in a
                       three-phase system, CALCULATE the following:
                       a.     Real power
                       b.     Reactive power
                       c.     Apparent power

         EO 1.7        Given a diagram of a wye- or delta-connected three-
                       phase system, DESCRIBE the voltage/current
                       relationships of the circuit.

         EO 1.8        STATE the indications of an unbalanced load in a three-
                       phase power system.


Three-Phase Systems

A three-phase (3φ) system is a combination of three single-phase systems. In a 3φ balanced
system, power comes from a 3φ AC generator that produces three separate and equal voltages,
each of which is 120° out of phase with the other voltages (Figure 10).




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                                     Figure 10 Three-Phase AC




Three-phase equipment (motors, transformers, etc.) weighs less than single-phase equipment of
the same power rating. They have a wide range of voltages and can be used for single-phase
loads. Three-phase equipment is smaller in size, weighs less, and is more efficient than
single-phase equipment.

Three-phase systems can be connected in two different ways. If the three common ends of each
phase are connected at a common point and the other three ends are connected to a 3φ line, it
is called a wye, or Y-, connection (Figure 11). If the three phases are connected in series to form
a closed loop, it is called a delta, or ∆-, connection.




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                               Figure 11 3φ AC Power Connections




Power in Balanced 3φ Loads

Balanced loads, in a 3φ system, have identical impedance in each secondary winding (Figure 12).
The impedance of each winding in a delta load is shown as Z∆ (Figure 12a), and the impedence
in a wye load is shown as Zy (Figure 12b). For either the delta or wye connection, the lines A,
B, and C supply a 3φ system of voltages.




                                  Figure 12 3φ Balanced Loads




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In a balanced delta load, the line voltage (VL) is equal to the phase voltage (Vφ ), and the line
current (IL) is equal to the square root of three times the phase current ( 3 Iφ ). Equation (9-5)
is a mathematical representation of VL in a balanced delta load. Equation (9-6) is a mathematical
representation of IL in a balanced delta load.

        VL = Vφ                                                                             (9-5)

        IL    3 Iφ                                                                          (9-6)

In a balanced wye load, the line voltage (VL) is equal to the square root of three times phase
voltage ( 3 Vφ ), and line current (IL) is equal to the phase current (Iφ ). Equation (9-7) is a
mathematical representation of VL in a balanced wye load. Equation (9-8) is a mathematical
representation of IL in a balanced wye load.

        VL        3 Vφ                                                                      (9-7)

        IL   Iφ                                                                             (9-8)

Because the impedance of each phase of a balanced delta or wye load has equal current, phase
power is one third of the total power. Equation (9-10) is the mathematical representation for
phase power (Pφ ) in a balanced delta or wye load.

        Pφ = Vφ Iφ cosθ                                                                    (9-10)

Total power (PT) is equal to three times the single-phase power. Equation (9-11) is the
mathematical representation for total power in a balanced delta or wye load.

        PT = 3Vφ Iφ cosθ                                                                   (9-11)

                                                 3 IL
In a delta-connected load, VL = Vφ and Iφ               so:
                                                 3

        PT        3 VL IL cos θ


                                               3 VL
In a wye-connected load, IL = Iφ and Vφ                 so:
                                                3

        PT        3 VL IL cos θ




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As you can see, the total power formulas for
delta- and wye-connected loads are identical.

Total apparent power (ST) in volt-amperes and
total reactive power (QT) in volt-amperes-reactive
are related to total real power (PT) in watts
(Figure 13).

A balanced three-phase load has the real,
apparent, and reactive powers given by:

         PT        3 VT IL cos θ                                 Figure 13 3φ Power Triangle


         ST        3 VT IL


         QT        3 VT IL sin θ

Example 1: Each phase of a delta-
connected 3φ AC generator
supplies a full load current of 200
A at 440 volts with a 0.6 lagging
power factor, as shown in Figure
14.

Find: 1.      VL
      2.      IL
      3.      PT
      4.      QT
      5.      ST


                                                     Figure 14 Three-Phase Delta Generator



Solution:

         1.        VL     Vφ

                   VL     440 volts




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        2.        IL    3 Iφ

                       (1.73)(200)

                  IL   346 amps


        3.        PT    3 VL IL cos θ

                       (1.73)(440)(346)(0.6)

                  PT   158.2 kW


        4.        QT     3 VL IL sin θ

                       (1.73)(440)(346)(0.8)

                  QT   210.7 kVAR


        5.        ST    3 VL IL

                       (1.73)(440)(346)

                  ST   263.4 kVA

Example 2: Each phase of a wye-
connected 3φ AC generator
supplies a 100 A current at a
phase voltage of 240V and a
power factor of 0.9 lagging, as
shown in Figure 15.

Find: 1.     VL
      2.     PT
      3.     QT
      4.     ST



                                                    Figure 15 Three-Phase Wye Generator




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Solution:

         1.   VL     3 Vφ

                   (1.73)(240)

              VL   415.2 volts


         2.   PT    3 VL IL cos θ

                   (1.73)(415.2)(100)(0.9)

              PT   64.6 kW


         3.   QT     3 VL IL sin θ

                   (1.73)(415.2)(100)(0.436)

              QT   31.3 kVAR


         4.   ST    3 VL IL

                   (1.73)(415.2)(100)

              ST   71.8 kVA

Unbalanced 3φ Loads

An important property of a three-phase balanced system is that the phasor sum of the three line
or phase voltages is zero, and the phasor sum of the three line or phase currents is zero. When
the three load impedances are not equal to one another, the phasor sums and the neutral current
(In) are not zero, and the load is, therefore, unbalanced. The imbalance occurs when an open or
short circuit appears at the load.

If a three-phase system has an unbalanced load and an unbalanced power source, the methods
of fixing the system are complex. Therefore, we will only consider an unbalanced load with a
balanced power source.


Example: A 3φ balanced system, as shown in Figure 16a, contains a wye load. The line-to- line
voltage is 240V, and the resistance is 40 Ω in each branch.



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                                      Figure 16 3φ Unbalanced Load



Find line current and neutral current for the following load conditions.
        1.     balanced load
        2.     open circuit phase A (Figure 16b)
        3.     short circuit in phase A (Figure 16c)


                                 Vφ               VL
        1.   IL     Iφ      Iφ             Vφ
                                 Rφ                 3

                    V 
                     L
                     
               IL    3
                     Rφ

                     240 
                          
                     1.73 
                       40

                    138.7
                     40

             IL     3.5 amps      IN      0



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         2.      Current flow in lines B and C becomes the resultant of the loads in B and C
                 connected in series.
                          VL
                 IB                   IC IB
                       RB RC

                         240
                       40 40

                 IB   3 amps               IC    3 amps

                 IN   IB         IC

                      3      3

                 IN   6 amps

                       VL
         3.      IB                   IC    IB
                       RB

                       240
                        40

                 IB   6 amps               IC    6 amps

                 The current in Phase A is equal to the neutral line current, IA = IN. Therefore, IN
                 is the phasor sum of IB and IC.
                 IN       3 IB

                      (1.73)(6)

                 IN   10.4 amps

In a fault condition, the neutral connection in a wye-connected load will carry more current than
the phase under a balanced load. Unbalanced three-phase circuits are indicated by abnormally
high currents in one or more of the phases. This may cause damage to equipment if the
imbalance is allowed to continue.




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Summary

Three-phase circuits are summarized below.


                           Three-Phase Circuits Summary

           Three-phase power systems are used in the industry because:

           -      Three-phase circuits weigh less than single-phase circuits of the same
                  power rating.
           -      They have a wide range of voltages and can be used for single-phase
                  loads.
           -      Three-phase equipment is smaller in size, weighs less, and is more
                  efficient than single-phase equipment.

           Unbalanced three-phase circuits are indicated by abnormally high currents in one
           or more of the phases.




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